(a, c, d)

We first assume that the discriminant is a perfect square. That is b 2 – 4ac = d 2 , d is an odd integer as a, b, c are odd.

Let a = 2k + 1, b = 2m + 1, c = 2n + 1, d = 2p + 1. Then (2m + 1) 2 – 4(2k + 1) (2n +1) = (2p +1) 2

⇒ (2m +1) 2 – (2p +1) 2 = 4(2k +1) (2n +1)

⇒ (2m – 2p) (2m + 2p + 1) = 4(2k +1) (2n +1) ⇒ (m – p) (m + p + 1) = (2k +1) (2n +1)

Now for any integral value of m and p, the L.H.S. of above is always even but R.H.S. is always odd, which is a contradiction. Hence, the discriminant cannot be perfect square. So the roots cannot be rational and hence cannot be integer also. Clearly discriminant cannot be zero.